Question

A metre stick is held vertically with one end on a rough horizontal floor. It is gently allowed to fall on the floor. Assuming that the end at the floor does not slip. find the angular speed of the rod when it hits the floor?

Answer 1

Let the mass of the rod $=m$

The potential energy of the rod in the vertical position will be at the center of mass of the rod.

The center of mass of the rod lies at its center.

$PE=Mg\frac{l}{2}$

On the other hand, as the rod falls, it gains rotational kinetic energy.

$E_R=\frac{1}{2}I{\omega }^2$

We know that the Moment of Inertia for a rod is $\frac{M{l}^2}{3}$

Therefore applying laws of conservation of energy

$\frac{1}{2}I{\omega }^2=mg\frac{l}{2}$

$\Rightarrow \frac{1}{2}\times \frac{M{l}^2}{3}\times {\omega }^2=mg\frac{l}{2}$

$\Rightarrow {\omega }^2=3g/l$

$\Rightarrow \omega =\sqrt{3g/l}$

put $g=9.8m/s^2$ and $l=1m$.

$=5.42rad/s$